Warm Up Explain 1) Why is an apple pies filling always hotter than the pastry even though they have been cooked at the same temperature? Your tongue is more likely to get burned by the filling than the crust

2 Reason Your tongue is more likely to get burned by the filling than the crust ( same with a hot pizza slice) pastry and filling have the same average kinetic energy ( temperature) as the pie comes out of the oven. 1)thermal conductivity pastry contains many pockets of air 2) specific heat capacity filling is mostly made of water, and water has a very high specific heat, the filling must give off a lot of heat for its temperature to decrease ( specific heat of water is 4.18 J/g°C ) ( specific heat to air is 1.01 J/g°C )

Don't use water to put out a grease fire oEOg oEOg

Q = mcΔT Q = mc (T f –T i ) i 2) What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? ( The specific heat capacity of water c = 4.18 J/g/°C. ) Show your work

Q = mCΔT What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? The specific heat capacity of water is 4.18 J/g/°C. Givens: m = 450 g c = 4.18 J/g/°C T initial = 15°C T final = 85°C Unknown Q = ??? ΔT = T final - T initial = 85°C - 15°C = 70.°C Q = mCΔT = (450 g)(4.18 J/g/°C)(70.°C) Q = J Q = 1.3x10 5

Relating the Quantity of Heat to the Temperature Change

3) What is change in temperature of 5 kg of water if it is given 84,000 J of heat energy? specific heat capacity of water = 4,200 J/(kg °C )

What is the rise in temperature of 5 kg of water if it is given 84,000 J of heat energy? specific heat capacity of water = 4,200 J/(kg K) Q = mcΔT Given: Q = 84,000 J m = 5 kg c = 4,200 J/(kg °C ) Looking for ΔT =? ΔT = Q / mc 84,000 J / ( 5kg x 4,200 J/kg °C) = 4 °C

4) What mass of water can be heated from 25.0° C to 50.0° C by the addition of 2,825 J?

Givens: c w = 4.18 J/g·K ∆T = 50.0° C – 25.0°C = 25.0 °C Unknown: m w = ? Formula Q = m w c w ∆T w m = Q/c ∆T Calculations m = (2825 J)/(4.18 J/g°C x 25.0°C) = 27.0 g

Calorimetry –is a technique to measure transfer of thermal energy ( heat flow )

Specific heat and the transfer of energy between two objects Use Conservation of energy Lost of Heat by an object = Gained in heat by the second objects | Q Lost | = | Q Gained | mc ( T i – T f ) = mc (T f –T i ) Final Temperature is the same ( the two objects are in thermal equilibrium)

Calorimetry – finding the specific heat of a 1kg horseshoe 5)A 1kg horseshoe has average kinetic energy of 200°C. The horseshoe is placed in 5 liter (5kg) bucket of water. Initially, the water has a temperature of 20°C.The horseshoe and the water reach a thermal equilibrium of 21.8 °C. Find the specific heat of the horseshoe

Given: Horseshoe m h = 1kg T ih = 200°C T fh = 21.8°C C h =???? Water m w = 5kg T iw = 20 °C T fw = 21.8°C c = 4,186 J/kg °C

Two ways to solve Isolate C h algebraically mc w ( T i – T f ) = m c h (T f –T i ) mc w ( T i – T f ) / (m h (T f –T i ))=c Q gained water = Q lost horseshoe Solve for Q Gained by water since we have all the variables. Q = mc w ( T i – T f ) Then c h = Q / ( m w ( T ih – T f ) J

Conservation of energy Lost of Heat by an object = Gained in heat by the second object | Q Lost | = | Q Gained | mc ( T i – T f ) = mc (T f –T i ) Final Temperature is the same ( the two objects are in thermal equilibrium) A 60 kg of aluminum that is at 15 °C is placed into 20 kg of 90 °C water. Find the final temperature Specific heat of aluminum is c aluminum = 900 J/ kg °C Specific heat of water c water = 4,186 J/kg °C )